Question: Find $\dfrac{d}{dx}\left[4x^2\cdot 2^{ (3x+1)}\right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $4x\cdot 2^{(3x+1)}\left(x+2\right)$ (Choice B) B $4x\cdot 2^{(3x+1)}\left(3x\ln(2)+2\right)$ (Choice C) C $\dfrac{8x\cdot 2^{(3x+1)}}{\ln(2)}$ (Choice D) D $24x\ln(2)\cdot 2^{(3x+1)}$
Solution: $4x^2\cdot 2^{ (3x+1)}$ is a product of a function and a composite function. Let... $u(x)=4x^2$ $v(x)=2^x$ $w(x)=3x+1$... then $4x^2\cdot 2^{ (3x+1)}=u(x)\cdot v\Bigl(w(x)\Bigr)$. To find $\dfrac{d}{dx}\left[4x^2\cdot 2^{ (3x+1)}\right]$, we will need to use the product rule and the chain rule! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u(x)\cdot v\Bigl(w(x)\Bigr)\right] \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \gray{\text{Product rule}} \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=8x$ $v'(x)=\ln(2)\cdot 2^x$ $w'(x)=3$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot{ v\Bigl(w(x)\Bigr)}+u(x)\cdot{ v'\Bigl(w(x)\Bigr)}\cdot w'(x) \\\\ &=8x\cdot{2^{(3x+1)}}+4x^2\cdot{\ln(2)\cdot 2^{ (3x+1)}}\cdot3 \\\\ &=8x\cdot 2^{(3x+1)}+12x^2\ln(2) \cdot 2^{(3x+1)} \\\\ &=4x\cdot 2^{(3x+1)}\left(2+3x\ln(2)\right) \\\\ &=4x\cdot 2^{(3x+1)}\left(3x\ln(2)+2\right) \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left[4x^2\cdot 2^{ (3x+1)}\right]= 4x\cdot 2^{(3x+1)}\left(3x\ln(2)+2\right)$.